1.
Expression                                               Value
----------                                               -----
(5 + 4 * 2 - 1 / 4) % 5                                  3
line.length() > 10 && numbers[1] > numbers.length        true
lineScan.hasNextInt() || (numbers[0] % 10 == 0)          false
3 * 3 + "TEST" + 1 + 1 + 2 * 2                           "9TEST114"
line.toUpperCase().charAt(numbers.length - numbers[2])   'R'


2.
Method Call                                Contents of array
-----------                                -----------------
int[] a1 = {7, 5};
mystery(a1);                               {1, 2}

int[] a2 = {3, 11, 4, 9};
mystery(a2);                               {1, 2, 1, 3}

int[] a3 = {0, 0, 0, 1, 1, 1};
mystery(a3);                               {1, 2, 3, 1, 2, 3}

int[] a4 = {26, 13, 37, -6, 16, 45, 12};
mystery(a4);                               {1, 1, 2, -6, 2, 3, 3}

int[] a5 = {-1, 1, -2, 16, 3, -3, -4, 5};
mystery(a5);                               {-1, 1, -2, 1, 2, -3, -4, 3}


3.
tuition
upperclass
ditch

tuition
books
newbie

party
old
study

tuition
books
study

4.
public static void bestGPA(Scanner input) {
    while (input.hasNextLine()) {
        String line = input.nextLine();
        Scanner lineScan = new Scanner(line);
        String bestName = lineScan.next();
        double bestGPA = lineScan.nextDouble();

        while (lineScan.hasNext()) {
            String name = lineScan.next();
            double gpa = lineScan.nextDouble();
            if (gpa > bestGPA) {
                bestGPA = gpa;
                bestName = name;
            }
        }
        System.out.println("Best GPA: " + bestGPA + " (" + bestName + ")");
    }
}


5.
public static int countDuplicates(int[] a) {
    int count = 0;
    for (int i = 1; i < a.length; i++) {
        if (a[i] == a[i - 1]) {
            count++;
        }
    }
    return count;
}


6. Two solutions are shown.

// 'count the lesser elements' solution
public static int[] ordinal(int[] a) {
    int[] result = new int[a.length];
    for (int i = 0; i < a.length; i++) {
        for (int j = 0; j < a.length; j++) {
            if (a[j] < a[i]) {
                result[i]++;
            }
        }
    }

    return result;
}

// 'sorted array' solution
public static int[] ordinal(int[] a) {
    int[] sorted = new int[a.length];
    for (int i = 0; i < a.length; i++) {
        sorted[i] = a[i];   // copy 'a' into 'sorted'
    }
    Arrays.sort(sorted);

    int[] result = new int[a.length];
    for (int i = 0; i < a.length; i++) {
        // index of element a[i] in sorted order should be stored as result[i]
        result[i] = Arrays.binarySearch(sorted, a[i]);
    }

    return result;
}


7. Two solutions are shown.

// 'count days in each month' solution
public int getDayOfYear() {
    int result = this.day;
    for (int i = 1; i < this.month; i++) {
        result += getDaysInMonth(i, this.year);
    }
    return result;
}

// 'count days from Jan. 1' solution
public int getDayOfYear() {
    Date temp = new Date(this.year, 1, 1);
    int result = 1;
    while (!this.equals(temp)) {
        temp.nextDay();
        result++;
    }
    return result;
}